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Pierce expansion of 2 - sqrt(3).
2

%I #25 Jan 05 2025 19:51:39

%S 3,5,51,53,140451,140453,2770663499604051,2770663499604053,

%T 21269209556953516583554114034636483645584976451,

%U 21269209556953516583554114034636483645584976453

%N Pierce expansion of 2 - sqrt(3).

%C For x in the open interval (0,1) define the map f(x) = 1 - x*floor(1/x). The n-th term (n >= 0) in the Pierce expansion of x is given by floor(1/f^(n)(x)), where f^(n)(x) denotes the n-th iterate of the map f, with the convention that f^(0)(x) = x.

%C The present sequence is the case x = 2 - sqrt(3).

%C Shallit has shown that the Pierce expansion of the quadratic irrational (c - sqrt(c^2 - 4))/2 has the form [c(0) - 1, c(0) + 1, c(1) - 1, c(1) + 1, c(2) - 1, c(2) + 1, ...], where c(0) = c and c(n+1) = c(n)^3 - 3*c(n). This is the case c = 4. For other cases see A006276 (c = 3), A219507 (c = 5) and A006275 (essentially c = 6 apart from the initial term).

%C The Pierce expansion of {(c - sqrt(c^2 - 4))/2}^(3^n) is [[c(n) - 1, c(n) + 1, c(n+1) - 1, c(n+1) + 1, c(n+2) - 1,c(n+2) + 1, ...].

%H G. C. Greubel, <a href="/A219506/b219506.txt">Table of n, a(n) for n = 0..13</a>

%H F. L. Bauer, <a href="http://www.ega-math.narod.ru/Nquant/Bauer.htm">Letters to the editor: An Infinite Product for Square-Rooting with Cubic Convergence</a>, The Mathematical Intelligencer, Vol. 20, Issue 1, (1998), 12-14.

%H T. A. Pierce, <a href="http://www.jstor.org/stable/2299963">On an algorithm and its use in approximating roots of algebraic equations</a>, Amer. Math. Monthly, Vol. 36 No. 10, (1929) p.523-525.

%H Jeffrey Shallit, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/22-4/shallit1.pdf">Some predictable Pierce expansions</a>, Fib. Quart., 22 (1984), 332-335.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PierceExpansion.html">Pierce Expansion</a>

%F a(2*n) = (2 + sqrt(3))^(3^n) + (2 - sqrt(3))^(3^n) - 1.

%F a(2*n + 1) = (2 + sqrt(3))^(3^n) + (2 - sqrt(3))^(3^n) + 1.

%F From _Peter Bala_, Jan 18 2022: (Start)

%F a(2*n+2) = a(2*n)^3 + 3*a(2*n)^2 - 3; a(2*n+1) = a(2*n-1)^3 - 3*a(2*n-1)^2 + 3.

%F a(2*n) = 6*(Product_{k = 1..n-1} a(2*k))^2 - 3, with a(0) = 1;

%F a(2*n+1) = 2*(Product_{k = 0..n-1} a(2*k+1))^2 + 3, with a(1) = 5.

%F sqrt(3) = (1 + 2/3)*(1 + 2/51)*(1 + 2/140451)*(1 + 2/2770663499604051)* .... See Bauer.

%F 1/sqrt(3) = (1 - 2/5)*(1 - 2/53)*(1 - 2/140453)*(1 - 2/2770663499604053)* .... (End)

%e We have the alternating series expansions

%e 2 - sqrt(3) = 1/3 - 1/(3*5) + 1/(3*5*51) - 1/(3*5*51*53) + ...

%e (2 - sqrt(3))^3 = 1/51 - 1/(51*53) + 1/(51*53*140451) - ...

%e (2 - sqrt(3))^9 = 1/140451 - 1/(140451*140453) + ....

%t PierceExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@ NestList[{Floor[1/Expand[1 - #[[1]] #[[2]]]], Expand[1 - #[[1]] #[[2]]]} &, {Floor[1/(A - Floor[A])], A - Floor[A]}, n - 1]]; PierceExp[N[2 - Sqrt[3] , 7!], 10] (* _G. C. Greubel_, Nov 14 2016 *)

%Y Cf. A006275, A006276, A112845, A219160, A219507, A219508.

%K nonn,easy

%O 0,1

%A _Peter Bala_, Nov 22 2012